x+6y=-6,x+3z-2y=16,2x+z+y=12.
如果|x+y+z-6|+|2x+3y-z-12|+|2x-y-z|=0求x,y,
1.x+y=16,y+z=12,z+x=102.3x-y+z=4,2x+3y-z=12,x+y+z=63.x+y+z=6
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
已知x、y、z满足方程组:x+y-z=6;y+z-x=2;z+x-y=0 求x、y、z的值
x+y+z=4 2x+3y-z=6 3x+2y+2z=10
2x-y+2z=-17 3x+y-3z=-4 x+y+z=6
2x+3y+z=11,3x+y-z=2,x+y+z=6求解
{2x+3y-4z=-5 x+y+z=6 x-y+3z=10
解方程组{3x+y-z=4,2x-y+3z=12,x+y+z=6}
{x+y+z=6,2x-y+z=3,3x+9y+z=24
解方程{3x -y+z=4 2x+3y-z=12 x+y+z=6
解方程组:x+y+z=4,x+y+2z=5,3x+y-z=6